Question

A buffer that contains 0.24 M of a base, B and 0.12 M of its conjugate acid BH⁺, has a pH of 9.5. What is the pH after 0.03 mol of NaOH are added to 0.67 L of the solution?

answer: 9.78

Answer 1

pOH = 14 - pH
= 14 - 9.5
= 4.5

use:
pOH = pKb + log {[BH+]/[B]}
4.5 = pKb + log (0.12/0.24)
4.5 = pKb -0.3
pKb = 4.8

before adding base:
mol of B = 0.24 M * 0.67 L = 0.1608 mol
mol of BH+ = 0.12 M * 0.67 L = 0.0804 mol

added 0.03 mol of NaOH
BH+ will react with NaOH to form B
after reaction,
mol of B = (0.1608 + 0.03) mol = 0.1908 mol
mol of BH+ = (0.0804 - 0.03) mol = 0.0504 mol

pOH = pKb + log {[BH+]/[B]}
= 4.8 + log (0.0504/0.1908)
= 4.8 - 0.58
= 4.22

pH = 14 - 4.22 = 9.78

Answer: 9.78